A Counterexample to a Corner-Peeling Conjecture for Subcomplexes of Z^3

Let G be a finite connected subgraph of the grid Z^3, not necessarily induced, and let X(G) be the cube complex whose cells are the squares and 3-cubes of Z^3 all of whose vertices and edges lie in G. A corner is a vertex that lies in exactly one maximal cell, and a corner peeling is an ordering v_1,...,v_n of the vertices in which each v_i is a corner of the subgraph induced by v_1,...,v_i. In the problem session of the 2026 Oberwolfach workshop Median Geometry and Applications, V. Chepoi, from discussions with J. Chalopin and M. Kokkou, conjectured that G admits a corner peeling whenever X(G) is simply connected and every connected component of every section of X(G) by a coordinate plane is simply connected. We show that the conjecture is false. We give a subgraph G_62 of Z^3 with 62 vertices in [0,3]^3 and an induced subgraph H_73 with 73 vertices in [0,4]^3 such that the cube complex and all its sections are collapsible, but there is no corner at all. Both examples have a symmetry group of order 6. The proofs can be checked by hand; explicit sequences of elementary collapses are supplied as machine-checkable certificates. A subdivision construction turns every finite subgraph into an induced subgraph whose pieces and corners are dilates of the original ones, so the partial- and induced-subgraph formulations are equivalent. Uncertified solver computations indicate that no counterexample fits into a 3 x 3 x 3 box of lattice points.